# RS Aggarwal Solutions Class 10 Ex 10D

SOLVE EACH OF THE FOLLOWING EQUATIONS BY USING THE METHOD OF COMPLETING THE SQUARE:

QUESTION-1:

$x^{2}-6x+3=0$

Solution:

$x^{2}-6x+3=0$

$\Rightarrow x^{2}-6x=-3$

$\Rightarrow x^{2}-2\times x\times 3+ 3^{2}=-3+3^{2}$ (Adding $3^{2}$ on both sides)

$\Rightarrow \left ( x-3 \right )^{2}=-3+9=6$

$\Rightarrow x-3 =\pm \sqrt{6}$ (Taking square root on both sides)

$\Rightarrow x-3 = \sqrt{6}\:or\:x-3=-\sqrt{6}$

$\Rightarrow x=3+\sqrt{6}\:or\:x=3-\sqrt{6}$

Hence, $3+\sqrt{6}\:and\:3-\sqrt{6}$ are the roots of the given equation.

QUESTION-2:

$x^{2}-4x+1=0$

Solution:

$x^{2}-4x+1=0$

=> $x^{2}-4x=-1$

=> $x^{2}-2\times x\times 2+2^{2}=-1+2^{2}$ (Adding $2^{2}$ on both sides)

=> $\left ( x-2 \right )^{2}=-1+4=3$

=> $\Rightarrow x-2 =\pm \sqrt{3}$ (Taking square root on both sides)

$\Rightarrow x-2 = \sqrt{3}\:or\:x-2=-\sqrt{3}$

$\Rightarrow x=2+\sqrt{3}\:or\:x=2-\sqrt{3}$

Hence, $2+\sqrt{3}\:and\:2-\sqrt{3}$ are the roots of the given equation.

QUESTION-3:

$x^{2}+8x-2=0$

Solution:

=> $x^{2}+8x=2$

=> $x^{2}+2\times x\times 4+4^{2}=2+4^{2}$ (Adding $4^{2}$ on both sides)

=> $\left ( x+4 \right )^{2}=2+16=18$

=> $x+4=\pm \sqrt{18}=\pm 3\sqrt{2}$ (Taking square root on both sides)

=> $x+4= 3\sqrt{2}\:or\:x+4=-3\sqrt{2}$

=> $x=-4+ 3\sqrt{2}\:or\:x=-4-3\sqrt{2}$

=> Hence, $\left ( -4+3\sqrt{2} \right )\:and\:\left ( -4-3\sqrt{2} \right )$ are the roots of the given equation.

QUESTION-4:

$4x^{2}+4\sqrt{3}x+3=0$

Solution:

$4x^{2}+4\sqrt{3}x+3=0$

=> $4x^{2}+4\sqrt{3}x=-3$

$\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \sqrt{3}+\left ( \sqrt{3} \right )^{2}$ (Adding $\left ( \sqrt{3} \right )^{2}$ on both sides)

$\Rightarrow \left ( 2x+\sqrt{3} \right )^{2}=-3+3=0$

$\Rightarrow 2x+\sqrt{3}=0$

$\Rightarrow x=-\frac{\sqrt{3}}{2}$

Hence, $-\frac{\sqrt{3}}{2}$ is the repeated root of the given equation.

QUESTION-5:

$2x^{2}+5x-3=0$

Solution:

$2x^{2}+5x-3=0$

$\Rightarrow 4x^{2}+10x-6=0$ (Multiplying both sides by 2)

$\Rightarrow 4x^{2}+10x=6$

$\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}$ [Adding $\left ( \frac{5}{2} \right )^{2}$ on both sides]

$\Rightarrow \left ( 2x+\frac{5}{2} \right )^{2}=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}=\left ( \frac{7}{2} \right )^{2}$

$\Rightarrow 2x+\frac{5}{2} =\pm \frac{7}{2}$ (Taking square root on both sides)

$\Rightarrow 2x+\frac{5}{2} = \frac{7}{2}\:or\:2x+\frac{5}{2}=-\frac{7}{2}$

$\Rightarrow 2x=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1\:or\:2x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6$

=> x = $\frac{1}{2}$ or x = -3

Hence, $\frac{1}{2}$ and -3 are the roots of the given equation.

QUESTION-6:

$3x^{2}-x-2=0$

Solution:

$3x^{2}-x-2=0$

$9x^{2}-3x-6=0$

$9x^{2}-3x=6$

$\Rightarrow \left ( 3x \right )^{2}-2\times 3x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=6+\left ( \frac{1}{2} \right )^{2}$ [Adding $\left ( \frac{1}{2} \right )^{2}$ on both sides]

$\Rightarrow \left ( 3x-\frac{1}{2} \right )^{2}=6+\frac{1}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}$

$\Rightarrow 3x-\frac{1}{2}=\pm \frac{5}{2}$ (Taking square root on both sides)

$\Rightarrow 3x-\frac{1}{2}=\frac{5}{2}\:or\:3x-\frac{1}{2}=-\frac{5}{2}$

$\Rightarrow 3x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\:or\:3x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$

=> x = 1 or x = $-\frac{2}{3}$

Hence, 1 and $-\frac{2}{3}$ are the roots of the given equation.

QUESTION-7:

$8x^{2}-14x-15=0$

Solution:

$8x^{2}-14x-15=0$

$\Rightarrow 16x^{2}-28x-30=0$ (Multiplying both sides by 2)

$\Rightarrow 16x^{2}-28x=30$

$\Rightarrow \left ( 4x \right )^{2}-2\times 4x\times \frac{7}{2}+\left ( \frac{7}{2} \right )^{2}=30+\left ( \frac{7}{2} \right )^{2}$ [Adding $\left ( \frac{7}{2} \right )^{2}$ on both sides]

$\Rightarrow \left ( 4x-\frac{7}{2} \right )^{2}=30+\frac{49}{4}=\frac{169}{4}=\left ( \frac{13}{2} \right )^{2}$

$\Rightarrow 4x-\frac{7}{2}=\pm \frac{13}{2}$ (Taking square root on both sides)

$\Rightarrow 4x-\frac{7}{2}=\frac{13}{2}\:or\:4x-\frac{7}{2}=-\frac{13}{2}$

$\Rightarrow 4x=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10\:or\:4x=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3$

=> x = $\frac{5}{2}$ or x = $-\frac{3}{4}$

Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.

QUESTION-8:

$7x^{2}+3x-4=0$

Solution:

$7x^{2}+3x-4=0$

$\Rightarrow 49x^{2}+21x-28=0$ (Multiplying both sides by 7)

$\Rightarrow 49x^{2}+21x=28$

$\Rightarrow \left ( 7x \right )^{2}+2\times 7x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=28+\left ( \frac{3}{2} \right )^{2}$ [Adding $\left ( \frac{3}{2} \right )^{2}$ on both sides]

$\Rightarrow \left ( 7x+\frac{3}{2} \right )^{2}=28+\frac{9}{4}=\frac{121}{4}=\left ( \frac{11}{2} \right )^{2}$

$\Rightarrow 7x+\frac{3}{2}=\pm \frac{11}{2}$ (Taking square root on both sides)

$\Rightarrow 7x+\frac{3}{2}=\frac{11}{2}\:or\:7x+\frac{3}{2}=-\frac{11}{2}$

$\Rightarrow 7x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4\:or\:7x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$

=> x = $\frac{4}{7}$ or x = -1

Hence, $\frac{4}{7}$ and -1 are the roots of the given equation.

QUESTION-9:

$3x^{2}-2x-1=0$

Solution:

$3x^{2}-2x-1=0$

$\Rightarrow 9x^{2}-6x-3=0$ (Multiplying both sides by 3)

$\Rightarrow 9x^{2}-6x=3$

$\Rightarrow \left ( 3x \right )^{2}-2\times 1+1^{2}=3+1^{2}$ [Adding $1^{2}$ on both sides]

$\Rightarrow \left (3x-1 \right)^{2}=3+1=4=\left ( 2 \right )^{2}$

$\Rightarrow 3x-1 =\pm 2$ (Taking square root on both sides)

=> 3x – 1 = 2 or 3x – 1 = -2

=> 3x = 3 or 3x = -1

=> x = 1 or x = $-\frac{1}{3}$

Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.

QUESTION-10:

$5x^{2}-6x-2=0$

Solution:

$5x^{2}-6x-2=0$

$\Rightarrow 25x^{2}-30x-10=0$ (Multiplying both sides by 5)

$\Rightarrow 25x^{2}-30x=10$

$\Rightarrow \left ( 5x \right )^{2}-2\times 5x\times3+3^{2}=10+3^{2}$ [Adding $3^{2}$ on both sides]

$\Rightarrow \left ( 5x-3 \right )^{2}=10+9=19$

$\Rightarrow 5x-3=\pm \sqrt{19}$ (Taking square root on both sides)

$\Rightarrow 5x-3= \sqrt{19}\:or\:5x-3=-\sqrt{19}$

$\Rightarrow 5x=3+\sqrt{19}\:or\:5x=3-\sqrt{19}$

$\Rightarrow x=\frac{3+\sqrt{19}}{5}\:or\:x=\frac{3-\sqrt{19}}{5}$

Hence, $\frac{3+\sqrt{19}}{5}\:and\:\frac{3-\sqrt{19}}{5}$ are the roots of the given equation.

QUESTION-11:

$\frac{2}{x^{2}}-\frac{5}{x}+2=0$

Solution:

$\Rightarrow \frac{2-5x+2x^{2}}{x^{2}}=0$

$\Rightarrow 2x^{2}-5x+2=0$

$\Rightarrow 4x^{2}-10x+4=0$ (Multiplying both sides by 2)

$\Rightarrow 4x^{2}-10x=-4$

$\Rightarrow \left ( 2x \right )^{2}-2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}=-4+\left ( \frac{5}{2} \right )^{2}$ [Adding $\left ( \frac{5}{2} \right )^{2}$ on both sides]

$\Rightarrow \left ( 2x-\frac{5}{2} \right )^{2}=-4+\frac{25}{4}=\frac{9}{4}=\left ( \frac{3}{2} \right )^{2}$

$\Rightarrow 2x-\frac{5}{2} = \pm \frac{3}{2}$ (Taking square root on both sides)

$\Rightarrow 2x-\frac{5}{2}=\frac{3}{2}\:or\:2x-\frac{5}{2}=-\frac{3}{2}$

$\Rightarrow 2x = \frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4\:or\:2x=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$

=> x = 2 or x = $\frac{1}{2}$

Hence, 2 and $\frac{1}{2}$ are the roots of the given equation.

QUESTION-12:

$4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0$

Solution:

$4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0$

$\Rightarrow 4x^{2}+4bx=a^{2}-b^{2}$

$\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times b+b^{2}=a^{2}-b^{2}+b^{2}$ [Adding $b^{2}$ on both sides]

$\Rightarrow \left ( 2x+b \right )^{2}=a^{2}$

$\Rightarrow 2x+b =\pm a$ (Taking square root on both sides)

=> 2x+b = a or 2x+b = -a

=> 2x = a-b or 2x = -a-b

$\Rightarrow x=\frac{a-b}{2}\:or\:x= -\frac{a+b}{2}$

Hence, $\frac{a-b}{2}\:and\:-\frac{a+b}{2}$ are the roots of the given equation.

Question-13:

$x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0$

Solution:

$x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0$

$x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=-\sqrt{2}$

$\Rightarrow x^{2}-2\times x\times \left ( \frac{\sqrt{2}+1}{2} \right )+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}=-\sqrt{2}+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}$ [Adding $\left ( \frac{\sqrt{2}+1}{2} \right )^{2}$ on both sides]

$\Rightarrow \left [ x-\left ( \frac{\sqrt{2}+1}{2}^{2} \right ) \right ]^{2}=\frac{-4\sqrt{2}+2+1+2\sqrt{2}}{4}=\frac{2-2\sqrt{2}+1}{4}=\left ( \frac{\sqrt{2}-1}{4} \right )^{2}$

$\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\pm \left ( \frac{\sqrt{2}-1}{2} \right )$ (Taking square root on both sides)

$\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\left ( \frac{\sqrt{2}-1}{2} \right )\:or\:x-\left ( \frac{\sqrt{2}+1}{2} \right )=-\left ( \frac{\sqrt{2}-1}{2} \right )$

$\Rightarrow x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\:or\:x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}$

$\Rightarrow x=\frac{2\sqrt{2}}{2}=\sqrt{2}\:or\:x=\frac{2}{2}=1$

Hence, $\sqrt{2}$ and 1 are the roots of the given equation.

Question-14:

$\sqrt{2}x^{2}-3x-2\sqrt{2}=0$

Solution:

$\sqrt{2}x^{2}-3x-2\sqrt{2}=0$

$\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x-4=0$ (Multiplying both sides by $\sqrt{2}$)

$\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x=4$

$\Rightarrow \left ( \sqrt{2}x \right )^{2}-2\times \sqrt{2}x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=4+\left ( \frac{3}{2} \right )^{2}$ (Adding $\left ( \frac{3}{2} \right )^{2}$ on both sides)

$\Rightarrow \left ( \sqrt{2}x-\frac{3}{2} \right )^{2}=4+\frac{9}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}$

$\Rightarrow \sqrt{2}x-\frac{3}{2}=\pm \frac{5}{2}$ (Taking square root on both sides)

$\Rightarrow \sqrt{2}x-\frac{3}{2}=\frac{5}{2}\:or\:\sqrt{2}x-\frac{3}{2}=-\frac{5}{2}$

$\Rightarrow \sqrt{2}x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\:or\:\sqrt{2}x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$

$\Rightarrow x=\frac{4}{\sqrt{2}}=2\sqrt{2}\:or\:x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$

Hence, $2\sqrt{2}\:and\:-\frac{\sqrt{2}}{2}$ are the roots of the given equation.

Question-15:

$\sqrt{3}x^{2}+10x+7\sqrt{3}=0$

Solution:

$\sqrt{3}x^{2}+10x+7\sqrt{3}=0$

$\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x+21=0$ (Multiplying both sides by $\sqrt{3}$)

$\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x=-21$

$\Rightarrow \left ( \sqrt{3x} \right )^{2}+2\times \sqrt{3}x\times 5+5^{2}=-21+5^{2}$ (Adding $5^{2}$ on both sides)

=> $\left ( \sqrt{3}x+5 \right )^{2}=-21+25=4=2^{2}$

=> $\sqrt{3}x+5=\pm 2$ (Taking square root on both sides)

=> $\sqrt{3}x+5=2\:or\:\sqrt{3}x+5=-2$

=> $\sqrt{3}x=-3\:or\:\sqrt{3}x=-7$

=> x = $-\frac{3}{\sqrt{3}}=-\sqrt{3}\:or\:x=-\frac{7}{\sqrt{3}}=-\frac{7\sqrt{3}}{3}$

Hence, $-3\:and\:-\frac{7\sqrt{3}}{3}$ are the roots of the given equation.

Question-16:

By using the method of completing the square, show that the equation $2x^{2}+x+4=0$ has no real roots.

Solution:

$2x^{2}+x+4=0$

=> $4x^{2}+2x+8=0$ (Multiplying both sides by 2)

=> $4x^{2}+2x=-8$

$\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=-8+\left ( \frac{1}{2} \right )^{2}$ (Adding $\left ( \frac{1}{2} \right )^{2}$ on both sides)

$\Rightarrow \left ( 2x+\frac{1}{2} \right )^{2}=-8+\frac{1}{4}=-\frac{31}{4}< 0$

But, $\left ( 2x+\frac{1}{2} \right )^{2}$ cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.

#### Practise This Question

If the people of india continue polluting the rivers at the same rate as of now. Then what are the worst things that can happen?